10
Parametric Equations and Polar
Coordinates
Copyright © Cengage Learning. All rights reserved.
10.5 Conic Sections
Copyright © Cengage Learning. All rights reserved.
Conic Sections
In this section we give geometric definitions of parabolas, ellipses, and
hyperbolas and derive their standard equations. They are called conic
sections, or conics, because they result from intersecting a cone with a plane
as shown in Figure 1.
Conics
Figure 1
3
Parabolas
4
Parabolas (1 of 9)
A parabola is the set of points in a plane that are equidistant from a fixed point
F (called the focus) and a fixed line (called the directrix). This definition is
illustrated by Figure 2.
Figure 2
5
Parabolas (2 of 9)
Notice that the point halfway between the focus and the directrix lies on the
parabola; it is called the vertex.
The line through the focus perpendicular to the directrix is called the axis of the
parabola.
In the 16th century Galileo showed that the path of a projectile that is shot into
the air at an angle to the ground is a parabola. Since then, parabolic shapes
have been used in designing automobile headlights, reflecting telescopes, and
suspension bridges.
6
Parabolas (3 of 9)
We obtain a particularly simple equation for a parabola if we place its vertex at
the origin O and its directrix parallel to the x-axis as in Figure 3.
Figure 3
7
Parabolas (4 of 9)
If the focus is the point (0, p), then the directrix has the equation y = −p. If P(x, y)
is any point on the parabola, then the distance from P to the focus is
PF 
x 2  ( y  p )2
and the distance from P to the directrix is y  p . (Figure 3 illustrates the case
where p > 0.)
8
Parabolas (5 of 9)
The defining property of a parabola is that these distances are equal:
x 2  ( y  p )2  y  p
We get an equivalent equation by squaring and simplifying:
2
x 2  ( y  p )2  y  p  ( y  p ) 2
x 2  y 2  2 py  p 2  y 2  2 py  p 2
x 2  4 py
9
Parabolas (6 of 9)
1 An equation of the parabola with focus (0, p) and directrix y = −p is
x 2  4 py
1
If we write a 
, then the standard equation of a parabola (1) becomes
 4p
2
y  ax .
10
Parabolas (7 of 9)
It opens upward if p > 0 and downward if p < 0 [see Figure 4, parts (a) and (b)].
(b) x 2  4 py , p  0
(a) x  4 py , p  0
2
Figure 4
11
Parabolas (8 of 9)
The graph is symmetric with respect to the y-axis because (1) is unchanged
when is replaced by −x.
If we interchange x and y in (1), we obtain the following.
2 An equation of the parabola with focus (p, 0) and directrix x = −p is
y 2  4 px
(Interchanging x and y amounts to reflecting about the diagonal line y = x.)
12
Parabolas (9 of 9)
The parabola opens to the right if p > 0 and to the left if p < 0 [see Figure 4,
parts (c) and (d)].
(d) y 2  4 px, p  0
(c) y 2  4 px, p  0
Figure 4
In both cases the graph is symmetric with respect to the x-axis, which is the
axis of the parabola.
13
Example 1
Find the focus and directrix of the parabola y 2  10 x  0 and sketch the graph.
Solution:
If we write the equation as y 2  10 x and compare it with Equation 2, we see
5
that 4p = −10, so p   .
2
5
 5

Thus the focus is  p, 0     , 0  and the directrix is x  .
2
 2 
14
Example 1 – Solution
The sketch is shown in Figure 5.
Figure 5
15
Ellipses
16
Ellipses (1 of 10)
An ellipse is the set of points in a plane the sum of whose distances from two
fixed points F1 and F2 is a constant (see Figure 6).
Figure 6
These two fixed points are called the foci (plural of focus). One of Kepler’s
laws is that the orbits of the planets in the solar system are ellipses with the sun
at one focus.
17
Ellipses (2 of 10)
In order to obtain the simplest equation for an ellipse, we place the foci on the
x-axis at the points (−c, 0) and (c, 0) as in Figure 7 so that the origin is halfway
between the foci.
P is on the ellipse when PF1  PF2  2a.
Figure 7
18
Ellipses (3 of 10)
Let the sum of the distances from a point on the ellipse to the foci be 2a > 0.
Then P(x, y) is a point on the ellipse when
PF1  PF2  2a
that is,
( x  c )2  y 2  ( x  c )2  y 2  2a
or
( x  c )2  y 2  2a  ( x  c )2  y 2
19
Ellipses (4 of 10)
Squaring both sides, we have
x 2  2cx  c 2  y 2  4a 2  4a ( x  c )2  y 2  x 2  2cx  c 2  y 2
which simplifies to
We square again:
which becomes
a ( x  c )2  y 2  a 2  cx
a 2 ( x 2  2cx  c 2  y 2 )  a 4  2a 2cx  c 2 x 2
(a 2  c 2 ) x 2  a 2 y 2  a 2 (a 2  c 2 )
20
Ellipses (5 of 10)
From triangle F1F2P in Figure 7 we see that 2c < 2a, so c < a and therefore
a 2  c 2  0. For convenience, let b 2  a 2  c 2 .
Then the equation of the ellipse becomes b 2 x 2  a 2 y 2  a 2 b 2 or, if both sides
are divided by a 2 b 2 ,
3
x2 y 2
 2 1
2
a
b
Since b 2  a 2  c 2  a 2 , it follows that b < a.
21
Ellipses (6 of 10)
x2
The x-intercepts are found by setting y = 0. Then 2  1, or x 2  a 2 , so x = ±a.
a
The corresponding points (a, 0) and (−a, 0) are called the vertices of the
ellipse and the line segment joining the vertices is called the major axis. To
find the y-intercepts we set x = 0 and obtain y 2  b 2 , so y = ±b.
The line segment joining (0, b) and (0, −b) is the minor axis.
22
Ellipses (7 of 10)
Equation 3 is unchanged if x is replaced by −x or y is replaced by −y, so the
ellipse is symmetric about both axes.
Notice that if the foci coincide, then c = 0, so a = b and the ellipse becomes a
circle with radius r = a = b.
23
Ellipses (8 of 10)
We summarize this discussion as follows (see also Figure 8).
4 The ellipse
x2 y 2
 2 1
2
a
b
ab0
has foci (±c, 0), where c 2  a 2  b 2 , and vertices (±a, 0).
x2 y 2

 1, a  b
a2 b2
Figure 8
24
Ellipses (9 of 10)
If the foci of an ellipse are located on the y-axis at (0, ±c), then we can find its
equation by interchanging x and y in (4). (See Figure 9.)
x2 y 2

 1, a  b
b2 a2
Figure 9
25
Ellipses (10 of 10)
5 The ellipse
x2 y 2
 2 1
2
b
a
ab0
has foci (0, ±c), where c 2  a 2  b 2 , and vertices (0, ±a).
26
Example 2
Sketch the graph of 9 x 2  16 y 2  144 and locate the foci.
Solution:
Divide both sides of the equation by 144:
x2 y 2

1
16 9
2
2
The equation is now in the standard form for an ellipse, so we have a  16, b  9,
a = 4, and b = 3.
The x-intercepts are ±4 and the y-intercepts are ±3.
27
Example 2 – Solution


Also, c 2  a 2  b 2  7, so c  7 and the foci are  7, 0 .
The graph is sketched in Figure 10.
9 x 2  16 y 2  144
Figure 10
28
Hyperbolas
29
Hyperbolas (1 of 6)
A hyperbola is the set of all points in a plane the difference of whose distances
from two fixed points F1 and F2 (the foci) is a constant. This definition is
illustrated in Figure 11.
P is on the hyperbola when PF1  PF2  2a.
Figure 11
30
Hyperbolas (2 of 6)
Notice that the definition of a hyperbola is similar to that of an ellipse; the only
change is that the sum of distances has become a difference of distances.
In fact, the derivation of the equation of a hyperbola is also similar to the one
given earlier for an ellipse.
When the foci are on the x-axis at (±c, 0) and the difference of distances is
PF1  PF2  2a, then the equation of the hyperbola is
6
x2 y 2
 2 1
2
a
b
where c 2  a 2  b 2 .
31
Hyperbolas (3 of 6)
Notice that the x-intercepts are again ±a and the points (a, 0) and (−a, 0) are
the vertices of the hyperbola.
But if we put x = 0 in Equation 6 we get y 2  b 2 , which is impossible, so there is
no y-intercept. The hyperbola is symmetric with respect to both axes.
To analyze the hyperbola further, we look at Equation 6 and obtain
x2
y2
 1 2  1
2
a
b
This shows that x 2  a 2 , so x 
x 2  a.
32
Hyperbolas (4 of 6)
Therefore we have x ≥ a or x ≤ −a. This means that the hyperbola consists of
two parts, called its branches.
When we draw a hyperbola it is useful to first draw its asymptotes, which are
the dashed lines
b
b
y    x and y     x shown in Figure 12.
a
a
x2 y 2

1
a2 b2
Figure 12
33
Hyperbolas (5 of 6)
Both branches of the hyperbola approach the asymptotes; that is, they come
arbitrarily close to the asymptotes.
7 The hyperbola
x2 y 2
 2 1
2
a
b
has foci (±c, 0), where c 2  a 2  b 2 , vertices (±a, 0), and asymptotes y    ab  x.
34
Hyperbolas (6 of 6)
If the foci of a hyperbola are on the y-axis, then by reversing the roles of x and
y we obtain the following information, which is illustrated in Figure 13.
8 The hyperbola
y 2 x2
 2 1
2
a
b
has foci (0, ±c), where c 2  a 2  b 2 ,
vertices (0, ±a), and asymptotes y    ab  x.
y 2 x2

1
a2 b2
Figure 13
35
Example 4
Find the foci and asymptotes of the hyperbola 9 x 2  16 y 2  144 and sketch its
graph.
Solution:
If we divide both sides of the equation by 144, it becomes
x2 y 2

1
16 9
which is of the form given in (7) with a = 4 and b = 3.
36
Example 4 – Solution
Since c 2  16  9  25, the foci are (±5, 0). The asymptotes are the lines
3
3
y  x and y   x. The graph is shown in Figure 14.
4
4
9 x 2  16 y 2  144
Figure 14
37
Shifted Conics
38
Shifted Conics (1 of 1)
We shift conics by taking the standard equations (1), (2), (4), (5), (7), and (8)
and replacing x and y by x − h and y − k.
39
Example 6
Find an equation of the ellipse with foci (2, −2), (4, −2) and vertices (1, −2), (5,
−2).
Solution:
The major axis is the line segment that joins the vertices (1, −2), (5, −2) and
has length 4, so a = 2.
The distance between the foci is 2, so c = 1.
Thus b 2  a 2  c 2  3.
40
Example 6 – Solution
Since the center of the ellipse is (3, −2), we replace x and y in (4) by x − 3 and
y + 2 to obtain
( x  3)2 ( y  2)2

1
4
3
as the equation of the ellipse.
41
Example 7
Sketch the conic 9 x 2  4 y 2  72 x  8 y  176  0 and find its foci.
Solution:
We complete the squares as follows:
4( y 2  2 y )  9( x 2  8 x )  176
4( y 2  2 y  1)  9( x 2  8 x  16)  176  4  144
4( y  1)2  9( x  4)2  36
42
Example 7 – Solution (1 of 2)
( y  1)2 ( x  4)2

1
9
4
This is in the form (8) except that x and y are replaced by x − 4 and y − 1. Thus
a 2  9, b 2  4, and c 2  13.
The hyperbola is shifted four units to the right and one unit upward.
43
Example 7 – Solution (2 of 2)
The foci are
 4, 1  13  and  4, 1  13  and the vertices are (4, 4) and (4, −2).
The asymptotes are y  1   32  x  4  . The hyperbola is sketched in Figure 15.
9 x 2  4 y 2  72 x  8 y  176  0
Figure 15
44